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Ta có: 4(5a + 3b) - 5(4a + 31b) = 143b => 5(4a + 31b) = 4(5a + 3b) - 143b
Vì 5a + 3b và 143b đều chia hết cho 13 nên 5(4a + 31b) chia hết cho 13. Mà (5;13) = 1
=> 4a + 31b chia hết cho 13
\(\dfrac{\left(12:13\right)+\left(12:131\right)-\left(12:1313\right)+\left(12:13131\right)}{\left(15:13\right)+\left(15:131\right)-\left(15:1313\right)+\left(15:13131\right)}\\ =\dfrac{12:\left(13+131-1313+13131\right)}{15:\left(13+131-1313+13131\right)}=\dfrac{12}{15}=\dfrac{4}{5}\)
a: \(A=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)
b: \(B=\dfrac{8+5}{10}:\dfrac{-5}{13}=\dfrac{13}{10}\cdot\dfrac{13}{-5}=-\dfrac{169}{100}\)
c: \(C=\left(\dfrac{88}{132}-\dfrac{33}{132}+\dfrac{60}{132}\right):\left(\dfrac{55}{132}+\dfrac{132}{132}-\dfrac{84}{132}\right)\)
\(=\dfrac{88-33+60}{55+132-84}=\dfrac{115}{103}\)
a: \(=\dfrac{-6}{11}:\dfrac{3\cdot11}{4\cdot5}=\dfrac{-6}{11}\cdot\dfrac{20}{33}=\dfrac{-2}{11}\cdot\dfrac{20}{11}=\dfrac{-40}{121}\)
b: \(=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)
c: \(=\dfrac{13}{10}:\dfrac{-5}{13}=\dfrac{-169}{50}\)
( - 12 ) . ( - 13 ) + 13 . ( - 29 )
= \(12.13+13.\left(-29\right)\)
=\(13.\left[12+\left(-29\right)\right]\)
=\(13.\left(-17\right)\)
=\(-221\)
a) (-12).(-13) + 13.(-29)
= 12.13 + 13.(-29)
= 13.(12 - 29)
= 13 . (-17)
=-221
b) -23.63 + 23.21 - 58.23
= -(23.63 - 23.21 + 58.23)
= -23.(63 - 21 + 58)
= -23.100
=-2300
a)
\(5+\left(-7\right)+9+\left(-11\right)+13+\left(-15\right)\)
\(=\left[5+\left(-7\right)\right]+\left[9+\left(-11\right)\right]+\left[13+\left(-15\right)\right]\)
\(=\left(-2\right)+\left(-2\right)+\left(-2\right)=-6\)
b)
\(\left(-6\right)+8+\left(-10\right)+12+\left(-14\right)+16\)
\(=\left[\left(-6\right)+8\right]+\left[\left(-10\right)+12\right]+\left[\left(-14\right)+16\right]\)
\(=2+2+2=6\)